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Find the set of solutions to this inequality?

$|x − 3| + |x − 6| < 5$

I have been taught to do it by treating $x$ in $3$ separate cases however I am not getting the correct answer. The answer is 'The set of real numbers $x$ such that $2 < x < 7.$ I am getting The set of real numbers $x$ such that $2 < x < 3$ or $6 < x < 7$.

Method

Case 1 $x<3$

$|x-3|= -(x-3)$

$|x-6|= -(x-6)$

$-(x-3)-(x-6)<5$

$-2x+9<5$

$-2x<-4$

$2x>4$

$x>2$

Because you have assumed $x<3$ the solutions is the intersection so $2<x<3$.

Case 2: $3<x<6$

$|x-3|= (x-3)$

$|x-6|= -(x-6)$

$(x-3)-(x-6)<5$

$3<5$

No solutions

Case 3: $x>6$

$|x-3|= (x-3)$

$|x-6|= (x-6)$

$(x-3)+(x-6)<5$

$2x-9<14$

$x<7$

Assumed $x>6$ therefore $6<x<7$.

So joint solutions the set of real numbers $x$ such that $2 < x < 3$ or $6 < x < 7$.

Where am I going wrong??

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Case 2:

3 < x < 6

|x-3|= (x-3)

|x-6|= -(x-6)

(x-3)-(x-6)<5

3<5

3<5 is true, so 3 < x < 6 is also solution for |x − 3| + |x − 6| < 5

Finally we have |x − 3| + |x − 6| < 5 $\Leftrightarrow$ 2 < x < 7

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